A solution of sucrose (molar mass 342 ${\mathrm{gmol}}^{-1}$) has been produced by dissolving 68.5g sucrose in 1000 g water. The freezing point of the solution obtained will be ${}^{\circ }\mathrm{C}$ $({\mathrm{K}}_{\mathrm{f}}$ for $\left.{\mathrm{H}}_2\mathrm{O}=1.86 \mathrm{kg}{ \mathrm{mol}}^{-1}\right)$
(Report in 3 decimals)

Depression in Freezing point $\Delta {\mathrm{T}}_{\mathrm{f}}=\frac{{\mathrm{W}}_1\times {\mathrm{K}}_{\mathrm{f}}\times 1000}{{\mathrm{M}}_1\times {\mathrm{W}}_2}$ where $W_1$= Weight of Solute
${\mathrm{W}}_2$= Weight of solvent
${\mathrm{M}}_1$= Molar mass of solute
${\mathrm{K}}_{\mathrm{f}}$= Freezing point deprssion constant
Now, $\Delta {\mathrm{T}}_{\mathrm{f}}=\frac{1.86\times 68.5\times 1000}{342\times 1000}=0.372\mathrm{C}$
Now, $\Delta {\mathrm{T}}_{\mathrm{f}}=\mathrm{T}°-{\mathrm{T}}_{\mathrm{f}}$
So, $T_f=0-0.372=-0.372C$.
(Freezing point of pure water $= 0°\mathrm{C}$.)

Hence the correct answer is -0.372⁰C.